Mechanics Exercise Class Ⅱ

1. Mechanics Exercise Class Ⅱ

2. Brief Review Work and Kinetic Energy Work Done by a Constant Force (Work-Kinetic Energy Theorem) Work Done by a Spring Force Work Done by a Variable Force Power

3. Work and Potential Energy Gravitational Potential Energy Elastic Potential Energy Conservation of Mechanical Energy The Center of Mass Conservation of Linear Momentum

4. 1. The force shown in the Fig. has magnitude Fp=20N and makes an angle of 300 to the ground .Calculate the work done by this force when the wagon is dragged 100m along the ground. Solution: We choose the x axis horizontal to the right and the y axis vertical upward. Then Whereas d=100m. Then using Eq

5. 2. A 5.0kg block moves in a straight line on a horizontal frictionless surface. Under the influence of a force that varies with position as shown in Fig. 7-31 How much work is done by the force as the block moves from the origin to x=8.0m? Solution: A block moves in a straight line, so Eq7-36 can be simplified as Key idea: the work done the system by the force component Fx as the system moves from xi to xf is the area under the curve between xi and xf . The work done is the area under the graph between x=0m to x=8.0m is

6. T A B o A B 3. A block of mass 0.5 kg moves horizontally along a frictionless rod attached to a cord of negligible mass. A force T of constant magnitude 50N acts on one end of the cord across a frictionless hoop. The block is moving toward the right side at point A. What is the velocity of the block when it is drawn to point B by force T? Solution: The work done by is zero. we get From Utilizing So

7. 2 O 1 h x 4. (P157.44)Two identical coins are initially held at height h=11.0 m. Coin 1 is dropped at time t=0 and then lands on a muddy field where it sticks. Coin 2 is dropped at t=0.500 s and then lands on the field. What is the acceleration of the center of mass (com) of the two-coin system (a) between t=0 and t=0.500 s, (b) between t=0.500 s and time t1 when coin hits and sticks, and (c) between t1 and time t2 when coin 2 hits and sticks? What is the speed of the center of mass when t is (d) 0.250 s, (e) 0.750 s, (f) 1.75s? Solution: Choose the dropping point as origin and downward as positive direction, construct x coordinate. (a) Between t=0 and t=0.500 s, coin 1 is free falling and coin 2 is at rest, so

8. (b) Between t=0.500 s and t1=1.50 s, coin 1 and 2 are both free falling, and (c) Between t1=1.50 s and t2=2.00s, coin 1 is at rest and coin 2 is free falling, and (d) When t=0.250 s, coin 1 is free falling and coin 2 is at rest, and the speed of com is (e) When t=0.750 s, 0.500s<t<t1, coin 1 and 2 are both free falling, and

9. (f) When t =1.75s, t1<t<t2, coin 1 is at rest and coin 2 is free falling, and

10. 5. (P154 21) In Fig.5-45, block 1 (mass 2.0 kg) is moving rightward at 10 m/s and block 2 (mass 5.0 kg) is moving rightward at 3.0 m/s. The surface is frictionless, and a spring with a spring constant of 1120 N/m is fixed to block 2. When the blocks collide, the compression of the spring is maximum at the instant the blocks have the same velocity. Find the maximum compression. Solution: The linear momentum of the two blocks and the spring system is conserved, and the mass of the spring is negligible. Choose rightward as positive direction and suppose when the compression of the spring is maximum, the velocity of the blocks is v, we have yields

11. Because of the compression of the spring, the total kinetic energy of the system decreased, gives us Substituting k=1120N/m into the above equation yields the maximum compression of the spring: